$ ext{Find the value of 'b' for which } (2x + 3) ext{ is a factor of } 2x^3 + 9x^2 - x - b ext{ :}$

Q: $\text{Find the value of 'b' for which } (2x + 3) \text{ is a factor of } 2x^3 + 9x^2 - x - b \text{ :}$

$\text{Find the value of 'b' for which } (2x + 3) \text{ is a factor of } 2x^3 + 9x^2 - x - b \text{ :}$

Question

$	ext{Find the value of 'b' for which } (2x + 3) 	ext{ is a factor of } 2x^3 + 9x^2 - x - b 	ext{ :}$

Accepted Answer

To determine the value of 'b' for which $ (2x + 3) $ is a factor of the polynomial $ 2x^3 + 9x^2 - x - b $, we can use the Factor Theorem. According to the Factor Theorem, if $ (2x + 3) $ is a factor of the polynomial, then substituting the root of $ 2x + 3 = 0 $ into the polynomial should yield zero.

First, find the root of $ 2x + 3 = 0 $:

$$
2x + 3 = 0 \implies 2x = -3 \implies x = -\frac{3}{2}
$$

Now, substitute $ x = -\frac{3}{2} $ into the polynomial $ 2x^3 + 9x^2 - x - b $ and set it equal to zero:

$$
2\left(-\frac{3}{2}\right)^3 + 9\left(-\frac{3}{2}\right)^2 - \left(-\frac{3}{2}\right) - b = 0
$$

Calculate each term:

1. $ 2\left(-\frac{3}{2}\right)^3 = 2 \times \left(-\frac{27}{8}\right) = -\frac{54}{8} = -\frac{27}{4} $

2. $ 9\left(-\frac{3}{2}\right)^2 = 9 \times \frac{9}{4} = \frac{81}{4} $

3. $ -\left(-\frac{3}{2}\right) = \frac{3}{2} $

Substitute these values back into the equation:

$$
-\frac{27}{4} + \frac{81}{4} + \frac{3}{2} - b = 0
$$

Combine the fractions:

$$
\left(\frac{81}{4} - \frac{27}{4}\right) + \frac{3}{2} - b = 0
$$

$$
\frac{54}{4} + \frac{3}{2} - b = 0
$$

Convert $\frac{3}{2}$ to a fraction with a denominator of 4:

$$
\frac{54}{4} + \frac{6}{4} - b = 0
$$

Combine the fractions:

$$
\frac{60}{4} - b = 0
$$

$$
15 - b = 0
$$

Solving for $ b $, we find:

$$
b = 15
$$

Thus, the correct value of 'b' is 15, which corresponds to Option 2.

**Explanation for Incorrect Options:**

1. **Option 1 (-25):** If $ b = -25 $, substituting into the polynomial would not satisfy the equation $ 2x^3 + 9x^2 - x - b = 0 $ when $ x = -\frac{3}{2} $.

2. **Option 3 (20):** Similarly, if $ b = 20 $, the substitution would not yield zero, indicating that $ (2x + 3) $ is not a factor.

3. **Option 4 (51):** If $ b = 51 $, the polynomial would not equal zero when evaluated at $ x = -\frac{3}{2} $, thus $ (2x + 3) $ would not be a factor.

Therefore, the only value of 'b' that makes $ (2x + 3) $ a factor of the polynomial is 15, confirming that Option 2 is correct.